The measure of each exterior angle is 60° and the polygon has 6 sides ⇒ 4th answer
Step-by-step explanation:
In a regular n-side polygon:
- All sides are equal in lengths
- All angles are equal in measure
- The measure of each interior angle =
- The measure of each exterior angle =
- The sum of interior angle and exterior angle at a vertex is 180°
∵ The measure of an interior angle of a regular polygon is 120°
∵ The sum of interior angle and exterior angle at a vertex = 180°
∴ The measure of each exterior angle = 180 - 120
∴ The measure of each exterior angle = 60°
∵ The measure of each exterior angle =
- Substitute the measure of the exterior angle by 60
∴ 
- By using cross multiplication
∴ 60 n = 360
- Divide both sides by 60
∴ n = 6
∴ The polygon has 6 sides
The measure of each exterior angle is 60° and the polygon has 6 sides
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Answer:
a) 240°
b) 30°
c) 225°
Step-by-step explanation:
To solve these equations you have to use the inverse of the given trigonometric functions. The inverse of <em>sin</em> is <em>arcsin</em>, and the inverse of <em>tan </em>is <em>arctan. </em>Instead of giving an angle, what is its sine?, the question is: given a sine, what is the angle?.
a)
sin(θ) = -√3/2
θ = arcsin(-√3/2)
θ = -60°
Given the periodicity of sine function, sin(-60°) is equivalent to sin(240°) (-60+180) and sin(300°) (-60+360).
b)
tan(θ) = 1/√3
θ = arctan(1/√3)
θ = 30°
c)
csc means cosecant, by definition:
csc(θ) = 1/sin(θ)
csc(θ) = -√2
1/sin(θ) = -√2
sin(θ) = -1/√2
θ = arcsin(-1/√2)
θ = -45° or 360-45 = 315° or 180+45 = 225°
Answer:
Not Parallel
Step-by-step explanation:
I think so sorry If I'm wrong
Rearrange the equation by writing
x² + x - 90 = 0
x² + x = 90
x² + x + 1/4 = 90 + 1/4
Recall that (a + b)² = a² + 2ab + b². Then
x² + 2 • x/2 + (1/2)² = 361/4
(x + 1/2)² = 361/4
x + 1/2 = ± √(361/4)
x = -1/2 ± 19/2
x = 9 or x = -10
Given the coordinates of the image of line segment RT to be R'(-2,-4) and T'(4.4), if the image produced was dilated by a scale factor of 12 centered at the origin, to get the coordinate of the end point, we will simply multiply the x and y coordinates of by the factor of 12 as shown:
For R' with coordinate R'(-2,-4), the coordinates of endpoint of the pre-image will be:
R = 12R'
R = 12(-2, -4)
R = (-24, -48)
For T' with coordinate T'(4,4), the coordinates of endpoint of the pre-imagee will be:
T = 12T'
T = 12(4, 4)
T = (48, 48)
Hence the coordinate of the endpoint of the preimage will be at R(-24, -48) and T(48, 48)
